Simplifying Trigonometric Identities

Question 1 of 7

Which of the following are trigonometric identities?

1. ${sec}^{2} θ = 1 + {tan}^{2} θ$ $\text(sec)^2 theta=1+\text(tan)^2 theta$
2. ${cot}^{2} θ = {cos}^{2} θ + 1$ $\text(cot)^2 theta=\text(cos)^2 theta+1$
3. ${csc}^{2} θ = {cot}^{2} θ + 1$ $\text(csc)^2 theta=\text(cot)^2 theta +1$
4. ${sec}^{2} θ = 1 + {sin}^{2} θ$ $\text(sec)^2 theta=1+\text(sin)^2 theta$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Using the given image, we can see that the radius of the circle will be

1

$1$ . Therefore, the circle follows the formula

x^{2} + y^{2} = 1

$x^2+y^2=1$ , where

1

$1$ is the radius.

Make a reference triangle by making a line from the point of origin

(0, 0)

$(0,0)$ going to any point in the circle to represent the radius and connect it to either the x or y-axis

Notice that, given angle

θ

$theta$ , the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent side

Now that we have the opposite and adjacent sides and the radius

1

$1$ as the hypotenuse, we can use the trigonometric functions to find their respective values

$sin$ $\text(sin)$	$=$ $=$	$\frac{opposite}{hypotenuse}$ $\text(opposite)/(\text(hypotenuse))$

	$=$ $=$	$\frac{y}{1}$ $y/1$

	$=$ $=$	$y$ $y$

$cos$ $\text(cos)$	$=$ $=$	$\frac{adjacent}{hypotenuse}$ $\text(adjacent)/(\text(hypotenuse))$

	$=$ $=$	$\frac{x}{1}$ $x/1$

	$=$ $=$	$x$ $x$

x

$x$ and

y

$y$ are both coordinate values, so we can write the coordinates as

(cos θ, sin θ)

$(\text(cos) theta,\text(sin) theta)$

Finally, use the values of

x

$x$ and

y

$y$ and derive its trigonometric identities

$x^{2} + y^{2}$ $x^2+y^2$	$=$ $=$	$1$ $1$
${cos}^{2} θ + {sin}^{2} θ$ $\text(cos)^2theta+\text(sin)^2theta$	$=$ $=$	$1$ $1$	Substitute known values
$({cos}^{2} θ + {sin}^{2} θ)$ $(\text(cos)^2theta+\text(sin)^2theta)$ $\div {cos}^{2} θ$ $divide \text(cos)^2theta$	$=$ $=$	$1$ $1$ $\div {cos}^{2} θ$ $divide \text(cos)^2theta$	Divide both sides by ${cos}^{2} θ$ $\text(cos)^2theta$

$1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $1+(\text(sin)^2theta)/(\text(cos)^2theta)$	$=$ $=$	$\frac{1}{{cos}^{2} θ}$ $1/(\text(cos)^2theta)$

$1 +$ $1+$ ${tan}^{2} θ$ $\text(tan)^2theta$	$=$ $=$	$\frac{1}{{cos}^{2} θ}$ $1/(\text(cos)^2theta)$	$\frac{{sin}^{2} θ}{{cos}^{2} θ} = {tan}^{2} θ$ $(\text(sin)^2theta)/(\text(cos)^2theta)=\text(tan)^2theta$

$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$	$\frac{1}{{cos}^{2} θ} = {sec}^{2} θ$ $1/(\text(cos)^2theta)=\text(sec)^2theta$
${sec}^{2} θ$ $\text(sec)^2theta$	$=$ $=$	$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$

$x^{2} + y^{2}$ $x^2+y^2$	$=$ $=$	$1$ $1$
${cos}^{2} θ + {sin}^{2} θ$ $\text(cos)^2theta+\text(sin)^2theta$	$=$ $=$	$1$ $1$	Substitute known values
$({cos}^{2} θ + {sin}^{2} θ)$ $(\text(cos)^2theta+\text(sin)^2theta)$ $\div {sin}^{2} θ$ $divide \text(sin)^2theta$	$=$ $=$	$1$ $1$ $\div {sin}^{2} θ$ $divide \text(sin)^2theta$	Divide both sides by ${sin}^{2} θ$ $\text(sin)^2theta$

$\frac{{cos}^{2} θ}{{sin}^{2} θ} + 1$ $(\text(cos)^2theta)/(\text(sin)^2theta)+1$	$=$ $=$	$\frac{1}{{sin}^{2} θ}$ $1/(\text(sin)^2theta)$

${cot}^{2} θ$ $\text(cot)^2theta$ $+ 1$ $+1$	$=$ $=$	$\frac{1}{{sin}^{2} θ}$ $1/(\text(sin)^2theta)$	$\frac{{cos}^{2} θ}{{sin}^{2} θ} = {cot}^{2} θ$ $(\text(cos)^2theta)/(\text(sin)^2theta)=\text(cot)^2theta$

${cot}^{2} θ + 1$ $\text(cot)^2theta+1$	$=$ $=$	${csc}^{2} θ$ $\text(csc)^2theta$	$\frac{1}{{sin}^{2} θ} = {csc}^{2} θ$ $1/(\text(sin)^2theta)=\text(csc)^2theta$
${csc}^{2} θ$ $\text(csc)^2theta$	$=$ $=$	${cot}^{2} θ + 1$ $\text(cot)^2theta+1$

{sec}^{2} θ = 1 + {tan}^{2} θ

$\text(sec)^2theta=1+\text(tan)^2theta$

{csc}^{2} θ = {cot}^{2} θ + 1

$\text(csc)^2theta=\text(cot)^2theta+1$

Question 2 of 7

Simplify

\frac{1}{1 + \sin θ} + \frac{1}{1 - \sin θ}

$1/(1+sin theta)+1/(1-sin theta)$

1. $2 {sec}^{2} θ$ $2\text(sec)^2theta$
2. $\frac{2}{{sec}^{2} θ}$ $2/(\text(sec)^2theta)$
3. ${csc}^{2} θ$ $\text(csc)^2theta$
4. ${csc}^{2} θ + 2$ $\text(csc)^2theta+2$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Apply the rule of adding fractions and simplify the expression

	$\frac{1}{1 + sin θ} + \frac{1}{1 - sin θ}$ $1/(1+\text(sin) theta)+1/(1-\text(sin) theta)$

$=$ $=$	$\frac{1 - sin θ + 1 + sin θ}{(1 + sin θ) (1 - sin θ)}$ $(1-\text(sin) theta+1+\text(sin) theta)/((1+\text(sin) theta)(1-\text(sin) theta))$

$=$ $=$	$\frac{2}{(1 + sin θ) (1 - sin θ)}$ $2/((1+\text(sin) theta)(1-\text(sin) theta))$	Combine like terms

$=$ $=$	$\frac{2}{1 - {sin}^{2} θ}$ $2/(1-\text(sin)^2theta)$	Expand

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$	$=$ $=$	$1$ $1$
${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	$=$ $=$	$1$ $1$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	Subtract ${sin}^{2} θ$ $\text(sin)^2theta$ from both sides
${cos}^{2} θ$ $\text(cos)^2theta$	$=$ $=$	$1 - {sin}^{2} θ$ $1-\text(sin)^2theta$

	$\frac{2}{1 - {sin}^{2} θ}$ $2/(1-\text(sin)^2theta)$

$=$ $=$	$\frac{2}{{cos}^{2} θ}$ $2/(\text(cos)^2theta)$	$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-\text(sin)^2theta=\text(cos)^2theta$

$=$ $=$	$\frac{2}{1} \times \frac{1}{{cos}^{2} θ}$ $2/1xx 1/(\text(cos)^2theta)$	Take out $2$ $2$

$=$ $=$	$2 \times {sec}^{2} θ$ $2xx\text(sec)^2theta$	$\frac{1}{{cos}^{2} θ} = {sec}^{2} θ$ $1/(\text(cos)^2theta)=\text(sec)^2theta$
$=$ $=$	$2 {sec}^{2} θ$ $2\text(sec)^2theta$

2 {sec}^{2} θ

$2\text(sec)^2theta$

Question 3 of 7

Simplify

\frac{\tan θ}{\sec θ}

$(tan theta)/(sec theta)$

1. $cos θ$ $\text(cos) theta$
2. $sin θ$ $\text(sin) theta$
3. $2 sec θ$ $2\text(sec) theta$
4. $\frac{1}{tan θ}$ $1/(\text(tan) theta)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Substitute known values to

tan

$\text(tan)$ and

sec

$\text(sec)$

$tan$ $\text(tan)$	$=$ $=$	$\frac{sin}{cos}$ $(\text(sin))/(\text(cos))$

$sec$ $\text(sec)$	$=$ $=$	$\frac{1}{cos}$ $1/(\text(cos))$

	$\frac{tan θ}{sec θ}$ $(\text(tan) theta)/(\text(sec) theta)$

$=$ $=$	$\frac{\frac{sin}{cos}}{\frac{1}{cos}}$ $((\text(sin))/(\text(cos)))/(1/(\text(cos)))$	Substitute known values

$=$ $=$	$sin θ$ $\text(sin) theta$	Apply rule of dividing fractions

sin θ

$\text(sin) theta$

Question 4 of 7

Simplify

\sec θ - \sec θ \sin^{2} θ

$sec theta-sec theta sin^2theta$

1. $sec θ$ $\text(sec) theta$
2. $csc θ$ $\text(csc) theta$
3. $cos θ$ $\text(cos) theta$
4. $sin θ$ $\text(sin) theta$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Factor out

sec θ

$\text(sec) theta$ on both terms

		$sec θ - sec θ {sin}^{2} θ$ $\text(sec) theta-\text(sec) theta \text(sin)^2theta$
	$=$ $=$	$sec θ (1 - {sin}^{2} θ)$ $\text(sec) theta(1-\text(sin)^2theta)$	Factorise

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$	$=$ $=$	$1$ $1$
${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	$=$ $=$	$1$ $1$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	Subtract ${sin}^{2} θ$ $\text(sin)^2theta$ from both sides
${cos}^{2} θ$ $\text(cos)^2theta$	$=$ $=$	$1 - {sin}^{2} θ$ $1-\text(sin)^2theta$

	$sec θ (1 - {sin}^{2} θ)$ $\text(sec) theta(1-\text(sin)^2theta)$
$=$ $=$	$sec θ ({cos}^{2} θ)$ $\text(sec) theta(\text(cos)^2theta)$	$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-\text(sin)^2theta=\text(cos)^2theta$

$=$ $=$	$\frac{1}{cos θ} \cdot ({cos}^{2} θ)$ $1/(\text(cos) theta)*(\text(cos)^2theta)$	$sec θ = \frac{1}{cos θ}$ $\text(sec) theta=1/(\text(cos) theta)$

$=$ $=$	$\frac{{cos}^{2} θ}{cos θ}$ $(\text(cos)^2theta)/(\text(cos) theta)$

$=$ $=$	$cos θ$ $\text(cos) theta$

cos θ

$\text(cos) theta$

Question 5 of 7

Simplify

\sec^{2} A - \csc^{2} A

$sec^2A-csc^2A$

1. $(sec A + csc A) (sec A + csc A)$ $(\text(sec) A+\text(csc) A)(\text(sec) A+\text(csc) A)$
2. $A (sec A + csc A)$ $A(\text(sec) A+\text(csc) A)$
3. $(sec A - csc A) (sec A + csc A)$ $(\text(sec) A-\text(csc) A)(\text(sec) A+\text(csc) A)$
4. $(sec A - csc A) (sec A - csc A)$ $(\text(sec) A-\text(csc) A)(\text(sec) A-\text(csc) A)$

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The difference of two squares is a format of a quadratic expression where there are two terms that are both perfect squares.

First, find the square of the two terms.

First term

{sec}^{2} A

$\text(sec)^2A$

=

$=$

sec A \cdot sec A

$\text(sec) A*\text(sec) A$

Second term

{csc}^{2} A

$\text(csc)^2A$

=

$=$

csc A \cdot csc A

$\text(csc) A*\text(csc) A$

Finally, get the sum and difference of the two values.

(

$($ $sec A$ $\text(sec) A$

-

$-$ $csc A$ $\text(csc) A$

) (

$)($ $sec A$ $\text(sec) A$

+

$+$ $csc A$ $\text(csc) A$

)

$)$

(sec A - csc A) (sec A + csc A)

$(\text(sec) A-\text(csc) A)(\text(sec) A+\text(csc) A)$

Question 6 of 7

Simplify

\frac{\tan^{2} θ}{\sec θ - 1}

$(tan^2theta)/(sec theta-1)$

1. $\frac{1}{sec θ}$ $1/(\text(sec)theta)$
2. $- \frac{1}{sec θ}$ $-1/(\text(sec)theta)$
3. $sec θ - 1$ $\text(sec)theta-1$
4. $sec θ + 1$ $\text(sec)theta+1$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Recall that

1 + {tan}^{2} θ = {sec}^{2} θ

$1+\text(tan)^2theta=\text(sec)^2theta$ . Derive this formula to further simplify the expression

$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$
$1 + {tan}^{2} θ$ $1+\text(tan)^2theta$ $- 1$ $-1$	$=$ $=$	${sec}^{2} θ$ $\text(sec)^2theta$ $- 1$ $-1$	Subtract $1$ $1$ from both sides
${tan}^{2} θ$ $\text(tan)^2theta$	$=$ $=$	${sec}^{2} θ - 1$ $\text(sec)^2theta-1$

	$\frac{{tan}^{2} θ}{sec θ - 1}$ $(\text(tan)^2theta)/(\text(sec) theta-1)$

$=$ $=$	$\frac{{sec}^{2} θ - 1}{sec θ - 1}$ $(\text(sec)^2theta-1)/(\text(sec) theta-1)$	${tan}^{2} θ = {sec}^{2} θ - 1$ $\text(tan)^2theta=\text(sec)^2theta-1$

$=$ $=$	$\frac{(sec θ - 1) (sec θ + 1)}{sec θ - 1}$ $((\text(sec)theta-1)(\text(sec)theta+1))/(\text(sec) theta-1)$	Factorise

$=$ $=$	$sec θ + 1$ $\text(sec)theta+1$	$\frac{sec θ - 1}{sec θ - 1} = 1$ $(\text(sec) theta-1)/(\text(sec) theta-1)=1$

sec θ + 1

$\text(sec)theta+1$

Question 7 of 7

Simplify

\frac{(\tan^{2} β) (\cot^{2} β + 1)}{\tan^{2} β + 1}

$((tan^2beta)(cot^2beta+1))/(tan^2beta+1)$

(1)

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Recall that

1 + {cot}^{2} θ = {csc}^{2} θ

$1+\text(cot)^2theta=\text(csc)^2theta$ . Derive this formula to further simplify the expression

	$\frac{({tan}^{2} β) ({cot}^{2} β + 1)}{{tan}^{2} β + 1}$ $((\text(tan)^2beta)(\text(cot)^2beta+1))/(\text(tan)^2beta+1)$

$=$ $=$	$\frac{{tan}^{2} β \cdot {csc}^{2} θ}{{tan}^{2} β + 1}$ $(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)$	$1 + {cot}^{2} θ = {csc}^{2} θ$ $1+\text(cot)^2theta=\text(csc)^2theta$

Finally, substitute known values to

tan

$\text(tan)$ ,

csc

$\text(csc)$ , and

sec

$\text(sec)$

$tan$ $\text(tan)$	$=$ $=$	$\frac{sin}{cos}$ $(\text(sin))/(\text(cos))$

$csc$ $\text(csc)$	$=$ $=$	$\frac{1}{sin}$ $1/(\text(sin))$

$sec$ $\text(sec)$	$=$ $=$	$\frac{1}{cos}$ $1/(\text(cos))$

	$\frac{{tan}^{2} β \cdot {csc}^{2} θ}{{tan}^{2} β + 1}$ $(\text(tan)^2beta*\text(csc)^2theta)/(\text(tan)^2beta+1)$

$=$ $=$	$\frac{\frac{{sin}^{2} β}{{cos}^{2} β} \cdot \frac{1}{{csc}^{2} θ}}{\frac{1}{{cos}^{2} θ}}$ $((\text(sin)^2beta)/(\text(cos)^2beta)*1/(\text(csc)^2theta))/(1/(\text(cos)^2theta))$	Substitute known values

$=$ $=$	$\frac{{sin}^{2} β}{{cos}^{2} β} \cdot \frac{1}{{csc}^{2} θ} \cdot \frac{{cos}^{2} θ}{1}$ $(\text(sin)^2beta)/(\text(cos)^2beta)1/(\text(csc)^2theta)(\text(cos)^2theta)/1$	Apply rule of dividing fractions

$=$ $=$	$1$ $1$

1

$1$

Simplifying Trigonometric Identities

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